### Application Of Derivatives Question 5

#### Question 5 - 29 January - Shift 2

Let $\alpha_1, \alpha_2, \ldots, \alpha_7$ be the roots of the equation $x^{7}+$ $3 x^{5}-13 x^{3}-15 x=0$ and $|\alpha_1| \geq|\alpha_2| \geq \ldots \geq|\alpha_7|$.

Then $\alpha_1 \alpha_2-\alpha_3 \alpha_4+\alpha_5 \alpha_6$ is equal to_______

## Show Answer

#### Answer: 9

#### Solution:

#### Formula: Roots of equations

Then $\alpha_1 \alpha_2-\alpha_3 \alpha_4+\alpha_5 \alpha_6$ is equal to

Given equation can be rearranged as

$ x(x^{6}+3 x^{4}-13 x^{2}-15)=0 $

clearly $x=0$ is one of the root and other part can

be observed by replacing $x^{2}=t$ from which we

have $\quad t^{3}+3 t^{2}-13 t-15=0$

$\Rightarrow$ $(t-3)(t^{2}+6 t+5)=0$

So, $\quad t=3, t=-1, t=-5$

Now we are getting $x^{2}=3, x^{2}=-1, x^{2}=-5$

$\Rightarrow x= \pm \sqrt{3}, x= \pm i, x= \pm \sqrt{5} i$

From the given condition $|\alpha_1| \geq|\alpha_2| \geq \ldots \geq|\alpha_7|$

We can clearly say that $\quad|\alpha_7|=0$ and

and

$|\alpha_6|=\sqrt{5}=|\alpha_5|$

and

$ |\alpha_4|=\sqrt{3}=|\alpha_3| \text{ and }|\alpha_2|=1=|\alpha_1| $

So we can have, $\alpha_1=\sqrt{5} i$, $\alpha_2=-\sqrt{5} i, \alpha_3=\sqrt{3} i$,

$\alpha_4=-\sqrt{3}, \alpha_5=i, \alpha_6=-i$

Hence

$ \begin{aligned} \alpha_1 \alpha_2-\alpha_3 \alpha_4 & +\alpha_5 \alpha_6 \\ & =1-(-3)+5=9 \end{aligned} $